Creating Database with PHP form

Dean_Glass

Hello all, I am a newbie to this PHP lark but I am finding it very interesting indeed. I am currently doing a project which involves creating and editing user CV details for online purposes. For some reason my test code is not working (how unusual!). I am trying to create a user in a pre-defined table in MySQL. My test connections are ok so I know that My configuration of MySQL and Apache are ok. (hopefully).

Here is the feeble snippet of code that does not work:

<html><head><title>NEW USER</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head><body>
<p><font size="4">Please enter username details:</font>
<?php

Ensuring that all fields have entries

if ($name and $pass)
{
#Connect to MySQL
$conn=@mysql_connect("localhost","root","")
or die("Sorry no connection");
#echo("MySQL connection test ok."."<br>");

#Select the specified database
$rs = @mysql_select_db("cvtest", $conn)
		or die("Database not located");
		#echo("Database located ok."."<br>");

#Create the query
$sql="insert into candidate (uname,password) values (\"$name\", \"$pass\" )";

#execute the query
$rs = @mysql_query($sql,$conn);

#confirm the added record details
if($rs) { echo("Record added: $name $pass");}

} #PROBLEMS WITH FORM NOT SUBMITTING VALUES TO SQL STATMENT SO NO DATA IS ENTERED!!
?>
</p>
<form action="<?php echo($PHP_SELF); ?>" method="post">
User Name: <input type="text" name="name" size="8">
Password: <input type="text" name="pass" size="8">
<br>
<br>
<input name="submit" type="submit" value="Create User">

</form></body></html>

If I comment out the "if"statement, then ther are two null entries in MySQL - Wierd! (but at least connection is ok). Something about the form is not passing on the variables "name" and "pass" so while I am checking to ensure that the fields are complete, no data gets passed to MySQL. I have a feeling in my water about the sql string but can't put my finger on it. If you are still awake after all this and can help, I would be very grateful indeed, as the rest of the project can follow this same format for all editing purposes. Thanx in advance.

jowatkins

I can't see any thing wrong with your query, the only thing I can suggest is to replace

$sql="insert into candidate (uname,password) values (\"$name\", \"$pass\" )";

with

$sql="insert into candidate (uname,password) values ('$name', '$pass' )";

By using single quotes, nothing will get confused. Also try echoing out your query to see what it's really reading.

Dean_Glass

Thank you for replying; unfortunately this did not work as the if statement contents are not executed on the ($name and $pass bit). Commenting these out gives me a table entry of null, null. Interestingly though, the echo for the sql staement is:

insert into candidate (uname,password) values ('', '' )

I might try to double-qoute, but I think I am on to a loser there, but as you can see something is not quite right. I shall keep scowering the "PHP in Easy Steps" book to see where I am going wrong.

MelbourneFL

well I'm new to php also so I hope you keep that in mind but ... since php is a serverside language, and since you're sending the form back to the server by calling the program itself, I don't think you're ever getting the actual fields back correctly. Here's what I use ...

at the beginning of the code use this to check for the fields:

if (isset ($HTTP_GET_VARS['user'])) {
$user = $HTTP_GET_VARS['user'];
}

if (isset ($HTTP_POST_VARS['pass'])) {
$pass = $HTTP_POST_VARS['pass'];
}

if ($user && $pass) {
......
}

then change your form header to :
<form action="<?php echo($PHP_SELF.'?user='.$user.'&pass='.$pass); ?>" method="post">

...

again, remember I'm just beginning in this language also so all I can do is tell you what's working for me 🙂 there are probably better ways to handle this and, most importantly, remember you're sending a password out in 'plain text' using this method.

HTH

Later,
Paul

Dean_Glass

Ooh, nice try. Did what you said but unfortunately the echoed sql string still is not receiving the data from the form (even though I can see the values in the address bar!). The echoed sql string is still

insert into candidate (uname,password) values ('','')

when I comment out the "If" statement. When the "if" statement is left in then the echoed command isn't reached 'cause the "if" criteria is never met. Differnet permutations have not worked with the sql statement, which have given me MySQL table entries from "," - "","" - Null,Null. I think the problem lays somewhere else, but I don't know where! There is definately (maybe) a parameter passing problem going on here somewhere - the $name and $pass are just not getting through to the sql staement from the form on submit.

Wakey, Wakey - I live in hope.

Dean_Glass

insert into candidate (uname,password) values ('','')

Wakey, Wakey - I live in hope.

bernouli

MelbourneFL was quite right about php, but the codes won't work because

if (isset ($HTTP_GET_VARS['user'])) {
$user = $HTTP_GET_VARS['user'];
}

if (isset ($HTTP_POST_VARS['pass'])) {
$pass = $HTTP_POST_VARS['pass'];
}

MelbourneFL uses GET method
since the form

<form action="<?php echo($PHP_SELF.'?user='.$user.'&pass='.$pass); ?>" method="post">

uses POST method

all to do is

if (isset ($HTTP_POST_VARS['user'])) {
$user = $HTTP_POST_VARS['user'];
}

if (isset ($HTTP_POST_VARS['pass'])) {
$pass = $HTTP_POST_VARS['pass'];
}

if ($user && $pass) {
......
}
<form action="<?php echo($PHP_SELF)?>" method="post">

Dean_Glass

Thanks, but it didnot work. However I have managed somehow to pass the password variable into the echoed reply - but this does not get its way into the MySQL table.

I thought I (erm..we/you) nearly had it then.

bernouli

well , trhy this, and put the error msg than here

 #Create the query
$sql="insert into candidate (uname,password) values ('".$user."', '".$pass."' )";

#execute the query
$rs = @mysql_query($sql,$conn) or die(mysql_error());

MelbourneFL

LOL sorry bout the GET vs POST thing ... anyway, I'm going to throw a guess here (and it's really dumb but ...) I put the variable name 'user' instead of 'name' ... I'm assuming you changed that first right? the only reason I ask is that you say you're getting the pass variable echoed but not the name ... one step at a time right?

I'm still thinking here

Later,
Paul

bernouli

auch, carefull with the $name and $user 🙂

Dean_Glass

Yeah, those were changed to my variables $name and $pass. New development though - Have managed to enter the password into MySQL, but only on the first run and with the "IF" parameters commented out.

Thanx anyway.

bernouli

which error did you get? no error?
try this if parameter

if ($HTTP_POST_VARS['name'] && $HTTP_POST_VARS['pass']) {
   $user = $HTTP_POST_VARS['name'];
   $pass = $HTTP_POST_VARS['pass'];
}

Dean_Glass

My error is Duplicate entry '' for key 3, so I gues that format is wrong maybe? - tryed your snippet but the error still remians. Still working on it. Thanx

MelbourneFL

null user name causing dup probably, still looking and thinking LOL

Later,
Paul

edit: try 'hard coding' a user name and let us know
edit: edit: key 3????? what's the table look like, you're only inserting two fields?

Dean_Glass

Yes of course, due to the first variable not being picked up and allways being null - that is causing the error. Doh!

Dean_Glass

Been hacking about a bit, but to no avail. It seems that if i keep the "if" staement in, then the form does nothing, if I keep it out then the defaults of null are entered. Here is my php file if anyone is interested, as I may be placing crucial elements in the wrong order.

Thanx in advance - Dean.

MelbourneFL

okay now you've confused me totally, I looked at the file and can't make heads or tails out of what you're doing now. You seem to have commented out all the suggestions???? It's probably just me but if you could post a "clean" copy of what's actually executing it will help I think.

Later,
Paul

Dean_Glass

Hard Coding seems to work fine - the connection and format of only putting in two fields in a fifteen field db seems ok as thier all optionally enterable. sql fragment:

$sql="insert into candidate (uname,password) values (\"Tae\",\"Kwondo\" )";

appear in MySQl fine when using the select statement to view them. The only interesting thing is that it only works when the "if" blocks are commented out, and then the hard coding names are its defaults. When the "if Statement blocks are back in then no code seems to be executed, The form clears ok but no data has passed or echoe produced.

Hmmm, a tough one for me I think - even though I am a Newbie.

Dean_Glass

Ok, sorry for any confusion but all the comments were for trial and error's ake as there were loads of good suggestions. Here is a clean (as you call it- well, cleanish) listing of new_user.php.